Question

Q.1) Find the area of the triangle whose sides are 13cm,14cm,15cm.

Answer 1

❈ Given ❈

→ A triangle with sides 13 cm, 14 cm and 15 cm.

Let us assume a triangle ABC with,

• AB = 13 cm
• BC = 14 cm
• CA = 15 cm

❈ To Find ❈

The area of triangle ABC.

❈ Solution ❈

$\bigstar\underline{\underline{\sf{DIAGRAM:-}}}$

$\setlength{\unitlength}{1.6mm}\begin{picture}(50,20)\linethickness{0.1mm}\put(-3,-3){\line(1,1){20}}\put(36.6,-2.8){\line(-1,1){19.8}}\put(-3,-3){\line(1,0){39.5}}\put(30,5){15 cm}\put(-1.5,5){13 cm}\put(3,5){}\put(15,-5){14 cm}\put(15.5,17.5){A}\put(-4,-5){B}\put(35,-5){C}\end{picture}$

Let us suppose :-

• a = 13 cm

• b = 14 cm

• c = 15 cm

→ Perimeter of the triangle = 2s = a + b + c

⇒ 2s = 13 + 14 + 15

⇒ s = 42/2

⇒ s = 21 cm

______________

s - a = 21 - 13

⇒ s - a = 8

______________

s - b = 21 - 14

⇒ s - b = 7

______________

s - b = 21 - 15

⇒ s - c = 6

______________

$\bigstar$ Now, applying Heron's formula :-

$\rm{A=\sqrt{s(s-a)(s-b)(s-c)} }\\\\\rm{\longrightarrow A= \sqrt{21(8)(7)(6)} }\\\\\rm{\longrightarrow A= \sqrt{3 \times 7 \times 7 \times 2 \times 2\times2\times2\times3} }\\\\\rm{\longrightarrow A= \sqrt{3^2\times7^2\times2^2\times2^2} }\\\\\rm{\longrightarrow A= 3\times7\times2\times2 }\\\\\rm{\longrightarrow A= 21\times4}\\\\\boxed{\boxed{\rm{\longrightarrow A= 84\ cm^2}}}$

Answer 2

Answer:

triangle with sides 13 cm, 14 cm and 15 cm.

Let us assume a triangle ABC with,

AB = 13 cm

BC = 14 cm

CA = 15 cm

❈ To Find ❈

The area of triangle ABC.

❈ Solution ❈

\bigstar\underline{\underline{\sf{DIAGRAM:-}}}★

DIAGRAM:−

\setlength{\unitlength}{1.6mm}\begin{picture}(50,20)\linethickness{0.1mm}\put(-3,-3){\line(1,1){20}}\put(36.6,-2.8){\line(-1,1){19.8}}\put(-3,-3){\line(1,0){39.5}}\put(30,5){15 cm}\put(-1.5,5){13 cm}\put(3,5){}\put(15,-5){14 cm}\put(15.5,17.5){A}\put(-4,-5){B}\put(35,-5){C}\end{picture}

Let us suppose :-

• a = 13 cm

• b = 14 cm

• c = 15 cm

→ Perimeter of the triangle = 2s = a + b + c

⇒ 2s = 13 + 14 + 15

⇒ s = 42/2

⇒ s = 21 cm

______________

s - a = 21 - 13

⇒ s - a = 8

______________

s - b = 21 - 14

⇒ s - b = 7

______________

s - b = 21 - 15

⇒ s - c = 6

______________

\bigstar★ Now, applying Heron's formula :-

\begin{gathered}\rm{A=\sqrt{s(s-a)(s-b)(s-c)} }\\\\\rm{\longrightarrow A= \sqrt{21(8)(7)(6)} }\\\\\rm{\longrightarrow A= \sqrt{3 \times 7 \times 7 \times 2 \times 2\times2\times2\times3} }\\\\\rm{\longrightarrow A= \sqrt{3^2\times7^2\times2^2\times2^2} }\\\\\rm{\longrightarrow A= 3\times7\times2\times2 }\\\\\rm{\longrightarrow A= 21\times4}\\\\\boxed{\boxed{\rm{\longrightarrow A= 84\ cm^2}}}\end{gathered}

A=

s(s−a)(s−b)(s−c)

⟶A=

21(8)(7)(6)

⟶A=

3×7×7×2×2×2×2×3

⟶A=

3

2

×7

2

×2

2

×2

2

⟶A=3×7×2×2

⟶A=21×4

⟶A=84 cm

2

Step-by-step explanation:

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